3

私はこのMySQLの問題の解決策を見つけるのに苦労しています。私はそれをどうやってやるのか頭を悩ませているようには思えません。次の表があります。

Question table
+----+-------------+
| id | question    |
+----+-------------+
| 1  | Is it this? |
| 2  | Or this?    |
| 3  | Or that?    |
+----+-------------+

Results Table
+----+---------+--------+
| id | user_id | job_id |
+----+---------+--------+
| 1  | 1       | 1      |
| 2  | 1       | 3      |
| 3  | 2       | 3      |
+----+---------+--------+

Answers table
+----+-------------------------+--------------+
| id | answer | fk_question_id | fk_result_id |
+----+-------------------------+--------------+
| 1  | Yes    | 1              | 1            |
| 2  | No     | 2              | 1            |
| 3  | Maybe  | 3              | 1            |
| 4  | Maybe  | 1              | 2            |
| 5  | No     | 2              | 2            |
| 6  | Maybe  | 3              | 2            |
| 7  | Yes    | 1              | 3            |
| 8  | Yes    | 2              | 3            |
| 9  | No     | 3              | 3            |
+----+-------------------------+--------------+

可能であれば、このように、質問の回答を各結果セットの列として表示したいと思います。

+-----------+---------+--------+-------------+----------+----------+
| result_id | user_id | job_id | Is it this? | Or this? | Or that? |
+-----------+---------+--------+-------------+----------+----------+
| 1         | 1       | 1      | Yes         | No       | Maybe    |
| 2         | 1       | 3      | Maybe       | No       | Maybe    |
| 3         | 2       | 3      | Yes         | Yes      | No       |
+-----------+---------+--------+-------------+----------+----------+

どんな助けでも大歓迎です。

ありがとう

4

1 に答える 1

3 
SELECT  a.ID,
        a.user_ID,
        a.job_id,
        MAX(CASE WHEN c.question = 'Is it this?' THEN b.answer END) 'Is it this?',
        MAX(CASE WHEN c.question = 'Or this?' THEN b.answer END) 'Or this?',
        MAX(CASE WHEN c.question = 'Or that? ' THEN b.answer END) 'Or that? '
FROM    Results a
        INNER JOIN Answers b
            ON a.id = b.fk_result_id
        INNER JOIN Question c
            ON b.fk_question_id = c.ID
GROUP   BY a.ID,
        a.user_ID,
        a.job_id

質問の数がわからない場合（具体的には、Matei Mihaiが言ったように1000）、動的バージョンが非常に必要です。

SET @sql = NULL;
SELECT
  GROUP_CONCAT(DISTINCT
    CONCAT(
      'MAX(CASE WHEN c.question = ''',
      question,
      ''' then b.answer end) AS ',
      CONCAT('`',question,'`')
    )
  ) INTO @sql
FROM Question;

SET @sql = CONCAT('SELECT  a.ID,
                            a.user_ID,
                            a.job_id, ', @sql, ' 
                    FROM    Results a
                            INNER JOIN Answers b
                                ON a.id = b.fk_result_id
                            INNER JOIN Question c
                                ON b.fk_question_id = c.ID
                    GROUP   BY a.ID,
                            a.user_ID,
                            a.job_id');

PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

出力

╔════╦═════════╦════════╦═════════════╦══════════╦══════════╗
║ ID ║ USER_ID ║ JOB_ID ║ IS IT THIS? ║ OR THIS? ║ OR THAT? ║
╠════╬═════════╬════════╬═════════════╬══════════╬══════════╣
║  1 ║       1 ║      1 ║ Yes         ║ No       ║ Maybe    ║
║  2 ║       1 ║      3 ║ Maybe       ║ No       ║ Maybe    ║
║  3 ║       2 ║      3 ║ Yes         ║ Yes      ║ No       ║
╚════╩═════════╩════════╩═════════════╩══════════╩══════════╝
于 2013-03-27T10:16:32.167 に答える